Q64 / 155 · WBJEE 2025 · Mathematics
The value of the integral $\int_0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$ is
- A$2$
- B$\frac{3}{4}$
- Correct answerC0
- D$-2$
Solution
$I=\int_0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$
Applying King's property
$I=\int_0^{\pi / 2} \log \left(\frac{4+3 \cos x}{4+3 \sin x}\right) d x$
$\begin{aligned} & 2 I=\int_0^{\pi / 2} \log \left(\left(\frac{4+3 \sin x}{4+3 \cos x}\right)\left(\frac{4+3 \cos x}{4+3 \sin x}\right)\right) d x=0 \\ & I=0\end{aligned}$
Hence, the correct answer is option: 'C'.